मान लीजिए $\frac{\pi}{2} < x < \pi$ इस प्रकार है कि $\cot x = \frac{-5}{\sqrt{11}}$ है। तो $\left(\sin \frac{11x}{2}\right)(\sin 6x - \cos 6x) + \left(\cos \frac{11x}{2}\right)(\sin 6x + \cos 6x)$ का मान ज्ञात कीजिए।

  • A
    $\frac{\sqrt{11}-1}{2\sqrt{3}}$
  • B
    $\frac{\sqrt{11}+1}{2\sqrt{3}}$
  • C
    $\frac{\sqrt{11}+1}{3\sqrt{2}}$
  • D
    $\frac{\sqrt{11}-1}{3\sqrt{2}}$

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यदि $\alpha, \beta$ न्यून कोण इस प्रकार हैं कि $\sin \beta=2 \sin \alpha$ और $3 \cos \beta=2 \cos \alpha$,तो $\sec (\alpha+\beta)=$

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